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Of course I am not implying that I am right and the no cloning theorem is wrong, but I am trying to figure out what is wrong with my reasoning and yet I couldn't find the mistake.

Based on Wikipedia

In physics, the no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.

So we start with a standard qubit $|\psi\rangle$ with completely unknown state where:

$$|\psi \rangle = \alpha |0\rangle + \beta|1\rangle$$

That qubit has probability $\alpha^2$ of being $0$ and probability $\beta^2$ of being 1 and if I understand the theory correctly it will not be possible to duplicate this qubit without knowing both $\alpha$ and $\beta$.

Now, if we plug this qubit along with a $|0\rangle$ into a $CNOT$ it seems to me that we end up with 2 identical qubits, each of them has probability $\alpha^2$ of being $0$ and probability $\beta^2$ of being 1.

Here is the math:

$$ CNOT |\psi, 0\rangle = \\ CNOT [(\alpha |0\rangle + \beta|1\rangle)\otimes |0\rangle] = \\ \begin{bmatrix}1&&0&&0&&0\\0&&1&&0&&0\\0&&0&&0&&1\\0&&0&&1&&0\end{bmatrix} \begin{pmatrix}\alpha\\\beta\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} =\\ \begin{bmatrix}1&&0&&0&&0\\0&&1&&0&&0\\0&&0&&0&&1\\0&&0&&1&&0\end{bmatrix} \begin{pmatrix}\alpha\\0\\\beta\\0\end{pmatrix} =\\ \begin{pmatrix}\alpha\\0\\0\\\beta\end{pmatrix} = \alpha \begin{pmatrix}1\\0\\0\\0\end{pmatrix}+\beta \begin{pmatrix}0\\0\\0\\1\end{pmatrix}=\\ \alpha |00\rangle + \beta |11\rangle $$ So the result becomes 2 exactly identical qubits, both have identical probabilities of being zero and identical probabilities of being one.

Since I am sure that the no-cloning theorem can't be wrong, I am asking what is wrong with my reasoning.

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    $\begingroup$ Unfortunately the no cloning theorem is often stated informally and "in words", the precise statement in my opinion is much easier to understand: given a state $|\psi\rangle$ and a blank state $|0\rangle$ independent of $|\psi\rangle$, there exists no unitary $U$ such that $U|\psi\rangle|0\rangle=e^{i\alpha}|\psi\rangle|\psi\rangle$ for some phase $\alpha$. This is clearly not what the CNOT does, as Mariia Mykhailova explains in her answer. $\endgroup$
    – user2723984
    Jul 31, 2019 at 9:07

1 Answer 1

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The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $\big(\alpha |0\rangle + \beta |1\rangle \big) \otimes \big(\alpha |0\rangle + \beta |1\rangle \big)$. This is not the state CNOT will give you.

The qubits you get after applying CNOT as you described do not, in fact, gave identical probabilities of measuring 0 and 1: as soon as you measure the first qubit, the measurement result of the second qubits will always the measurement result of the first qubit! If you were able to actually clone the qubit, the measurement results of the second qubit would not depend on the results of measuring the first one.

(You can also browse other questions about no-cloing theorem to find different explanations.)

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